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11.E: Electrochemical Methods (Exercises)

1. Identify the anode and cathode for the following electrochemical cells, and write the oxidation or reduction reaction at each electrode.

a. \(\mathrm{Pt \:|\: FeCl_2(\mathit{aq},0.015),FeCl_3(\mathit{aq},0.045) \:||\: AgNO_3(\mathit{aq},0.1) \:|\: Ag}\)

b. \(\mathrm{Ag \:|\: AgBr(\mathit{s}),NaBr(\mathit{aq},1.0) \:||\: CdCl_2(\mathit{aq},0.05) \:|\: Cd}\)

c. \(\mathrm{Pb \:|\: PbSO_4(\mathit{s}),H_2SO_4(\mathit{aq},1.5) \:||\: H_2SO_4(\mathit{aq},2.0),PbSO_4(\mathit{s}) \:|\: PbO_2}\)

2. Calculate the potential for the electrochemical cells in problem 1. The values in parentheses are the activities of the associated species.

3. Calculate the activity of KI, x, in the following electrochemical cell if the potential is +0.294 V

\(\mathrm{Ag \:|\: AgCl(\mathit{s}),NaCl(\mathit{aq},0.1) \:||\: KI(\mathit{aq,x}),I_2(\mathit{s}) \:|\: Pt}\)

4. What reaction prevents us from using Zn as an electrode of the first kind in acidic solutions? Which other electrodes of the first kind would you expect to behave in the same manner as Zn when immersed in an acidic solution?

5. Creager and colleagues designed a salicylate ion-selective electrode using a PVC membrane impregnated with tetraalkylammonium salicylate.20 To determine the ion-selective electrode’s selectivity coefficient for benzoate, they prepared a set of salicylate calibration standards in which the concentration of benzoate was held constant at 0.10 M. Using the following data, determine the value of the selectivity coefficient.

[salicylate] (M)

potential (mV)

1.0

20.2

1.0×10–1

73.5

1.0×10–2

126

1.0×10–3

168

1.0×10–4

182

1.0×10–5

182

1.0×10–6

177

What is the maximum acceptable concentration of benzoate if you plan to use this ion-selective electrode to analyze samples containing as little as 10–5 M salicylate with an accuracy of better than 1%?

6. Watanabe and co-workers described a new membrane electrode for the determination of cocaine, a weak base alkaloid with a pKa of 8.64.21 The electrode’s response for a fixed concentration of cocaine is independent of pH in the range of 1–8, but decreases sharply above a pH of 8. Offer an explanation for this pH dependency.

7. Figure 11.20 shows a schematic diagram for an enzyme electrode that responds to urea by using a gas-sensing NH3 electrode to measure the amount of ammonia released following the enzyme’s reaction with urea. In turn, the NH3 electrode uses a pH electrode to monitor the change in pH due to the ammonia. The response of the urea electrode is given by equation 11.14. Beginning with equation 11.11, which gives the potential of a pH electrode, show that equation 11.14 for the urea electrode is correct.

8. Explain why the response of an NH3-based urea electrode (Figure 11.20 and equation 11.14) is different from the response of a urea electrode in which the enzyme is coated on the glass membrane of a pH electrode (Figure 11.21 and equation 11.15).

9. A potentiometric electrode for HCN uses a gas-permeable membrane, a buffered internal solution of 0.01 M KAg(CN)2, and a Ag2S ISE electrode that is immersed in the internal solution. Consider the equilibrium reactions taking place within the internal solution, and derive an equation relating the electrode’s potential to the concentration of HCN in the sample.

10. Mifflin and associates described a membrane electrode for the quantitative analysis of penicillin in which the enzyme penicillinase is immobilized in a polyacrylamide gel coated on the glass membrane of a pH electrode.22 The following data were collected using a set of penicillin standards.

[penicillin] (M)

potential (mV)

1.0×10–2

220

2.0×10–3

204

1.0×10–3

190

2.0×10–4

153

1.0×10–4

135

1.0×10–5

96

1.0×10–6

80

(a) Over what range of concentrations is there a linear response?

(b) What is calibration curve’s equation for this concentration range?

(c) What is the concentration of penicillin in a sample that yields a potential of 142 mV?

11. An ion-selective electrode can be placed in a flow cell into which we inject samples or standards. As the analyte passes through the cell, a potential spike is recorded instead of a steady-state potential. The concentration of K+ in serum has been determined in this fashion using standards prepared in a matrix of 0.014 M NaCl.23

[K+] (mM) potential (arb. units)
0.1 25.5
0.2 37.2
0.4 50.8
0.6 58.7
0.8 64.0
1.0 66.8

A 1.00-mL sample of serum is diluted to volume in a 10-mL volumetric flask and analyzed, giving a potential of 51.1 (arbitrary units). Report the concentration of K+ in the sample of serum.

12. Wang and Taha described an interesting application of potentiometry, which they call batch injection.24 As shown in Figure 11.56, an ionselective electrode is placed in an inverted position in a large volume tank, and a fixed volume of a sample or a standard solution is injected toward the electrode’s surface using a micropipet. The response of the electrode is a spike in potential that is proportional to the analyte’s concentration. The following data were collected using a pH electrode and a set of pH standards.

pH potential (mV)
2.0 +300
3.0 +240
4.0 +168
5.0 +81
6.0 +35
8.0 –92
9.0 –168
10.0 –235
11.0 –279

Determine the pH of the following samples given the recorded peak potentials: tomato juice, 167 mV; tap water, –27 mV; coffee, 122 mV.

Figure11.56.jpg

Figure 11.56 Schematic diagram for a batch injection analysis. See Problem 11.12 for more details.

13. The concentration of NO3 in a water sample is determined by a onepoint standard addition using a NO3 ion-selective electrode. A 25.00-mL sample is placed in a beaker and a potential of 0.102 V is measured. A 1.00-mL aliquot of a 200.0-mg/L standard solution of NO3 is added, after which the potential is 0.089 V. Report the mg NO3/L in thewater sample.

14. In 1977, when I was an undergraduate student at Knox College, my lab partner and I did an experiment to determine the concentration of fluoride in tap water and the amount of fluoride in toothpaste. The data in this problem comes from my lab notebook.

(a) To analyze tap water, we took three 25.0-mL samples and added 25.0 mL of TISAB to each. We measured the potential of each solution using a F ISE and an SCE reference electrode. Next, we made five 1.00-mL additions of a standard solution of 100.0 ppm F to each sample, and measure the potential after each addition.

mL of
standard added

sample 1
potential (mV)
sample 2

sample 3
0.00 –79 –82 –81
1.00 –119 –119 –118
2.00 – 133 – 133 – 133
3.00 – 142 – 142 – 142
4.00 – 149 – 148 – 148
5.00 – 154 – 153 – 153

Report the parts-per-million of F in the tap water.

(b) To analyze the toothpaste, we measured 0.3619 g into a 100-mL volumetric flask, added 50.0 mL of TISAB, and diluted to volume with distilled water. After ensuring that the sample was thoroughly mixed, we transferred three 20.0-mL portions into separate beakers and measured the potential of each using a F ISE and an SCE reference electrode. Next, we made five 1.00-mL additions of a standard solution of 100.0 ppm F to each sample, and measured the potential after each addition.

Note

For a more thorough description of this analysis, see Representative Method 11.1.

mL of
standard added


sample 1

potential (mV)
sample 2


sample 3

0.00 – 55 – 54 – 55
1.00 – 82 – 82 – 83
2.00 – 94 – 94 – 94
3.00 – 102 – 103 – 102
4.00 – 108 – 108 – 109
5.00 – 112 – 112 – 113

Report the parts-per-million F in the toothpaste.

15. You are responsible for determining the amount of KI in iodized salt and decide to use an I ion-selective electrode. Describe how you would perform this analysis using external standards and using the method of standard additions.

16. Explain why each of the following decreases the analysis time for controlled-potential coulometry: a larger surface area for the working electrode, a smaller volume of solution, and a faster stirring rate.

17. The purity of a sample of picric acid, C6H3N3O7, is determined by controlled-potential coulometry, converting the picric acid to triaminophenol, C6H9N3O.

picric acid to triaminophenol.png

A 0.2917-g sample of picric acid is placed in a 1000-mL volumetric flask and diluted to volume. A 10.00-mL portion of this solution is transferred to a coulometric cell and sufficient water added so that the Pt cathode is immersed. The exhaustive electrolysis of the sample requires 21.67 C of charge. Report the purity of the picric acid.

18. The concentration of H2S in the drainage from an abandoned mine can be determined by a coulometric titration using KI as a mediator and I3 as the titrant.

\[\ce{H2S}(aq) + \ce{I3-}(aq) + \ce{2H2O}(l) ⇋ \ce{2H3O+}(aq) + \ce{3I-}(aq) + \ce{S}(s)\]

A 50.00-mL sample of water is placed in a coulometric cell, along with an excess of KI and a small amount of starch as an indicator. Electrolysis is carried out at a constant current of 84.6 mA, requiring 386 s to reach the starch end point. Report the concentration of H2S in the sample in parts-per-million.

19. One method for the determination of H3AsO3 is a coulometric titration using I3 as a titrant. The relevant standard-state reactions and potentials are summarized here.

\[\ce{H3AsO4}(aq) + \ce{2H3O+}(aq) + 2e^− ⇋ \ce{H3AsO3}(aq) + \ce{3H2O}(l) \hspace{20px} E^\circ = \mathrm{+0.559\: V}\]

\[\ce{I3-}(aq) + 2e^− ⇋ 3\ce{I-}(aq) \hspace{20px} E^\circ = \mathrm{+0.536\: V}\]

Explain why the coulometric titration must be carried out in a neutral solution (pH ≈ 7) instead of in a strongly acidic solution (pH < 0).

20. The production of adiponitrile, NC(CH2)4CN, from acrylonitrile, CH2=CHCN, is an important industrial process. A 0.594-g sample of acrylonitrile is placed in a 1-L volumetric flask and diluted to volume. An exhaustive controlled-potential electrolysis of a 1.00-mL portion of the diluted acrylonitrile requires 1.080 C of charge. What is the value of n for the reduction of acrylonitrile to adiponitrile?

21. The linear-potential scan hydrodynamic voltammogram for a mixture of Fe2+ and Fe3+ is shown in Figure 11.57, where il,a and il,c are the anodic and cathodic limiting currents.

(a) Show that the potential is given by

\[E = E^\circ_\mathrm{Fe^{3+}/Fe^{2+}} - 0.05916\log\dfrac{K_\mathrm{Fe^{3+}}}{K_\mathrm{Fe^{2+}}} − 0.05916\log\dfrac{i - i_\textrm{l,a}}{i_\textrm{l,c} - i}\]

(b) What is the potential when i = 0 for a solution that is 0.1 mM Fe3+ and 0.05 mM Fe2+? You may assume that KFe3+KFe2+.

Figure11.57.jpg

Figure 11.57 Linear-scan hydrodynamic voltammogram for a mixture of Fe2+ and Fe3+. See Problem 11.21 for more details.

22. The amount of sulfur in aromatic monomers can be determined by differential pulse polarography. Standard solutions are prepared for analysis by dissolving 1.000 mL of the purified monomer in 25.00 mL of an electrolytic solvent, adding a known amount of S, deaerating, and measuring the peak current. The following results were obtained for a set of calibration standards.

μg S added peak current (μA)

0

0.14

28

0.70

56

1.23

112

2.41

168

3.42

Analysis of a 1.000-mL sample, treated in the same manner as the standards, gives a peak current of 1.77 μA. Report the mg S/mL in the sample.

23. The purity of a sample of K3Fe(CN)6 was determined using linear-potential scan hydrodynamic voltammetry at a glassy carbon electrode. The following data were obtained for a set of external calibration standards.

[K3Fe(CN)6] (mM)

limiting current (μA)

2.0

127

4.0

252

6.0

376

8.0

500

10.0

624

A sample of impure K3Fe(CN)6 was prepared for analysis by diluting a 0.246-g portion to volume in a 100-mL volumetric flask. The limiting current for the sample was found to be 444 μA. Report the purity of this sample of K3Fe(CN)6.

24. One method for determining whether an individual has recently fired a gun is to look for traces of antimony in the residue collected from the individual’s hands. Anodic stripping voltammetry at a mercury film electrode is ideally suited for this analysis. In a typical analysis a sample is collected from a suspect with a cotton-tipped swab wetted with 5% v/v HNO3. After returning to the lab, the swab is placed in a vial containing 5.0 mL of 4 M HCl that is 0.02 M in hydrazine sulfate. After allowing the swab to soak overnight, a 4.0-mL portion of the solution is transferred to an electrochemical cell along with 100 mL of 0.01 M HgCl2. After depositing the thin film of mercury and the antimony, the stripping step gives a peak current of 0.38 μA. After adding a standard addition of 100 μL of 5.00×102 ppb Sb, the peak current increases to 1.14 μA. How many nanograms of Sb were collected from the suspect’s hand?

25. Zinc can be used as an internal standard in the analysis of thallium by differential pulse polarography. A standard solution containing 5.00 × 10–5 M Zn2+ and 2.50× 10–5 M Tl+ gave peak currents of 5.71 μA and 3.19 μA, respectively. An 8.713-g sample of an alloy known to be free of zinc was dissolved in acid, transferred to a 500-mL volumetric flask, and diluted to volume. A 25.0-mL portion of this solution was mixed with 25.0 mL of a 5.00 × 10–4 M solution of Zn2+. Analysis of this solution gave a peak current for Zn2+ of 12.3 μA, and for Tl+ of 20.2 μA. Report the %w/w Tl in the alloy.

26. Differential pulse voltammetry at a carbon working electrode can be used to determine the concentrations of ascorbic acid and caffeine in drug formulations.25 In a typical analysis a 0.9183-g tablet is crushed and ground into a fine powder. A 0.5630-g sample of this powder is transferred to a 100-mL volumetric flask, brought into solution, and diluted to volume. A 0.500-mL portion is then transferred to a voltammetric cell containing 20.00 mL of a suitable supporting electrolyte. The resulting voltammogram gives peak currents of 1.40 μA and 3.88 μA for ascorbic acid and caffeine, respectively. A 0.500-mL aliquot of a standard solution containing 250.0 ppm ascorbic acid and 200.0 ppm caffeine is then added. A voltammogram of this solution gives peak currents of 2.80 μA and 8.02 μA for ascorbic acid and caffeine, respectively. Report the milligrams of ascorbic acid and milligrams of caffeine in the tablet.

27. Ratana-ohpas and co-workers described a stripping analysis method for determining the amount of tin in canned fruit juices.26 Standards containing 50.0 ppb Sn4+, 100.0 ppb Sn4+, and 150.0 ppb Sn4+ were analyzed giving peak currents (arbitrary units) of 83.0, 171.6, and 260.2, respectively. A 2.00-mL sample of lychee juice was mixed with 20.00 mL of 1:1 HCl/HNO3. A 0.500-mL portion of this mixture was added to 10 mL of 6 M HCl and the volume adjusted to 30.00 mL. Analysis of this diluted sample gave a signal of 128.2 (arbitrary units). Report the parts-per-million Sn4+ in the original sample of lychee juice.

28. Sittampalam and Wilson described the preparation and use of an amperometric sensor for glucose.27 The sensor is calibrated by measuring the steady-state current when it is immersed in standard solutions of glucose. A typical set of calibration data is shown here.

[glucose] (mg/100 mL) current (arb. units)

2.0

17.2

4.0

32.9

6.0

52.1

8.0

68.0

10.0

85.8

A 2.00-mL sample is diluted to 10 mL in a volumetric flask and a steady-state current of 23.6 (arbitrary units) is measured. What is the concentration of glucose in the sample in mg/100 mL?

29. Differential pulse polarography can be used to determine the concentrations of lead, thallium, and indium in a mixture. Because the peaks for lead and thallium, and for thallium and indium overlap, a simultaneous analysis is necessary. Peak currents (in arbitrary units) at –0.385 V, –0.455 V, and –0.557 V were measured for a single standard solution, and for a sample, giving the results shown in the following table.

standards peak currents (arb. units) at
analyte μg/mL –0.385 V –0.455 V –0.557 V
Pb2+ 1.0 26.1 2.9 0
Tl+ 2.0 7.8 23.5 3.2
In3+ 0.4 0 0 22.9
sample 60.6 28.8 54.1

Report the μg/mL of Pb2+, Tl+ and In3+ in the sample.

30. Abass and co-workers developed an amperometric biosensor for NH4+ that uses the enzyme glutamate dehydrogenase to catalyze the following reaction

\[\textrm{2-oxyglutarate}(aq) + \ce{NH4+}(aq) + \ce{NADH}(aq) → \ce{glutamate}(aq) + \ce{NAD+}(aq) + \ce{H2O}(l)\]

where NADH is the reduced form of nicotinamide adenine dinucleotide.28 The biosensor actually responds to the concentration of NADH, however, the rate of the reaction depends on the concentration of NH4+. If the initial concentrations of 2-oxyglutarate and NADH are the same for all samples and standards, then the signal is proportional to the concentration of NH4+. As shown in the following table, the sensitivity of the method is dependent on pH.

pH sensitivity (nA s–1 M–1)
6.2 1.67 × 103
6.75 5.00× 103
7.3 9.33 × 103
7.7 1.04 × 104
8.3 1.27 × 104
9.3 2.67 × 103

Two possible explanations for the effect of pH on the sensitivity of this analysis are the acid–base chemistry of NH4+, and the acid–base chemistry of the enzyme. Given that the pKa for NH4+ is 9.244, explain the source of this pH-dependent sensitivity.

31. The speciation scheme for trace metals shown in Table 11.12 divides them into seven operationally defined groups by collecting and analyzing two samples following each of four treatments—a total of eight samples and eight measurements. After removing insoluble particulates by filtration (treatment 1), the solution is analyzed for the concentration of ASV labile metals and for the total concentration of metals. A portion of the filtered solution is passed through an ion-exchange column (treatment 2), and the concentrations of ASV metal and of total metal are determined. A second portion of the filtered solution is irradiated with UV light (treatment 3), and the concentrations of ASV metal and of total metal are measured. Finally, a third portion of the filtered solution is irradiated with UV light and passed through an ion-exchange column (treatment 4), and the concentrations of ASV labile metal and of total metal again are determined. The groups that are included in each measurement are summarized in the following table.


treatment

groups removed
by treatment

groups contributing to
ASV-labile metals

groups contributing to
total metals

1

none

I, II, III

I, II, III, IV, V,

VI, VII

2

I, IV, V

II, III

II, III, VI, VII

3

none

I, II, III, IV, VI

I, II, III, IV, V,

VI, VII

4

I, II, IV, V, VI

III

III, VII

(a) Explain how you can use these eight measurements to determine the concentration of metals present in each of the seven groups identified in Table 11.12.

(b) Batley and Florence report the following results for the speciation of cadmium, lead, and copper in a sample of seawater.29

measurement

(treatment: ASV-labile or total)

 

ppm Cd2+

 

ppm Pb2+

 

ppm Cu2+

1: ASV-labile

0.24

0.39

0.26

1: total

0.28

0.50

0.40

2: ASV-labile

0.21

0.33

0.17

2: total

0.26

0.43

0.24

3: ASV-labile

0.26

0.37

0.33

3: total

0.28

0.50

0.43

4: ASV-labile

0.00

0.00

0.00

4: total

0.02

0.12

0.10

Determine the speciation of each metal in this sample of sea water.

32. The concentration of Cu2+ in seawater may be determined by anodic stripping voltammetry at a hanging mercury drop electrode after first releasing any copper bound to organic matter. To a 20.00-mL sample of seawater is added 1 mL of 0.05 M HNO3 and 1 mL of 0.1% H2O2. The sample is irradiated with UV light for 8 hr and then diluted to volume in a 25-mL volumetric flask. Deposition of Cu2+ takes place at –0.3 V versus an SCE for 10 min, producing a peak current of 26.1 (arbitrary units). A second 20.00-mL sample of the seawater is treated identically, except that 0.1 mL of a 5.00 μM solution of Cu2+ is added, producing a peak current of 38.4 (arbitrary units). Report the concentration Cu2+ in the seawater in mg/L.

33. Thioamide drugs can be determined by cathodic stripping analysis.30 Deposition occurs at +0.05 V versus an SCE. During the stripping step the potential is scanned cathodically, and a stripping peak is observed at –0.52 V. In a typical application a 2.00-mL sample of urine was mixed with 2.00 mL of a pH 4.78 buffer. Following a 2.00 min deposition, a peak current of 0.562 μA was measured. A 0.10-mL addition of a 5.00 μM solution of the drug was added to the same solution. A peak current of 0.837 mA was recorded using the same deposition and stripping conditions. Report the drug’s molar concentration in the urine sample.

34. The concentration of vanadium (V) in sea water can be determined by adsorptive stripping voltammetry after forming a complex with catechol.31 The catechol-V(V) complex is deposited on a hanging mercury drop electrode at a potential of –0.1 V versus a Ag/AgCl reference electrode. A cathodic potential scan gives a stripping peak that is proportional to the concentration of V(V). The following standard additions were used to analyze a sample of seawater.

[V(V)]added (M)

peak current (nA)

2.0×10–8

24

4.0×10–8

33

8.0×10–8

52

1.2×10–7

69

1.8×10–7

97

2.8×10–7

140

Determine the molar concentration of V (V) in the sample of sea water, assuming that the standard additions result in a negligible change in the sample’s volume.

35. The standard-state reduction potential for Cu2+ to Cu is +0.342 V versus the SHE. Given that Cu2+ forms a very stable complex with the ligand EDTA, do you expect the standard-state reduction potential for Cu(EDTA)2– to be greater than +0.342 V, less than +0.342 V, or equal to +0.342 V? Explain your reasoning.

36. The polarographic half-wave potentials (versus the SCE) for Pb2+ and Tl+ in 1 M HCl are, respectively, –0.44 V and –0.45 V. In an electrolyte of 1 M NaOH, however, the half-wave potentials are –0.76 V for Pb2+ and –0.48 V for Tl+. Why does the change in electrolyte have such a significant effect on the half-wave potential for Pb2+, but not on the half-wave potential for Tl+?

37. The following data for the reduction of Pb2+ was collected by normal-pulse polarography.

potential (V vs. SCE)

current (μA)     

–0.345

0.16

–0.370

0.98

–0.383

2.05

–0.393

3.13

–0.409

4.62

–0.420

5.16

The limiting current was 5.67 μA. Verify that the reduction reaction is reversible, and determine values for n and E1/2. The half-wave potentials for the normal-pulse polarograms of Pb2+ in the presence of several different concentrations of OH are shown in the following table.

[OH] (M)

E1/2 (V vs. SCE)

[OH] (M)

E1/2 (V vs. SCE)

0.050

–0.646

0.150

–0.689

0.100

–0.673

0.300

–0.715

Determine the stoichiometry of the Pb-hydroxide complex and its formation constant.

38. In 1977, when I was an undergraduate student at Knox College, my lab partner and I did an experiment to study the voltammetric behavior of Cd2+ (in 0.1 M KNO3) and Ni2+ (in 0.2 M KNO3) at a dropping mercury electrode. The data in this problem comes from my lab notebook. All potentials are relative to an SCE reference electrode.

potential for Cd2+ (V)

current (μA)

–0.60

4.5

–0.58

3.4

–0.56

2.1

–0.54

0.6

–0.52

0.2

 

potential for Ni2+ (V)

current (μA)

–1.09

1.90

–1.05

1.75

–1.03

1.50

–1.02

1.25

–1.01

1.00

The limiting currents for Cd2+ was 4.8 μA and that for Ni2+ was 2.0 μA. Evaluate the electrochemical reversibility for each metal ion and comment on your results.

39. Baldwin and co-workers report the following data from a cyclic voltammetry study of the electrochemical behavior of p-phenylenediamine in a pH 7 buffer.32 All potentials are measured relative to an SCE.

scan rate (mV/s)

Ep,a (V)

Ep,c (V)

ip,a (mA)

ip,c (mA)

2

0.148

0.104

0.34

0.30

5

0.149

0.098

0.56

0.53

10

0.152

0.095

1.00

0.94

20

0.161

0.095

1.44

1.44

50

0.167

0.082

2.12

1.81

100

0.180

0.063

2.50

2.19

The initial scan is toward more positive potentials, leading to the oxidation reaction shown here.

p-phenylenediamine ox.png

Use this data to show that the reaction is electrochemically irreversible. A reaction may show electrochemical irreversibility because of slow electron transfer kinetics or because the product of the oxidation reaction participates in a chemical reaction that produces an nonelectroactive species. Based on the data in this problem, what is the likely source of p-phenylenediamine’s electrochemical irreversibility?

11.5.3 Solutions to Practice Exercises

Practice Exercise 11.1

The oxidation of H2 to H+ occurs at the anode

\[\ce{H2}(g) ⇋ \ce{2H+}(aq) + 2e^-\]

and the reduction of Cu2+ to Cu occurs at the cathode.

\[\ce{Cu^2+}(aq) + 2e^- ⇋ \ce{Cu}(s)\]

The overall cell reaction, therefore, is

\[\ce{Cu^2+}(aq) + \ce{H2}(g) ⇋ \ce{2H+}(aq) + \ce{Cu}(s)\]

Click here to return to the chapter.

Practice Exercise 11.2

Making appropriate substitutions into equation 11.3 and solving for Ecell gives its value as

\[\begin{align}
E_\ce{cell} &= \left(E^\circ_\mathrm{Cu^{2+}/Cu} – \dfrac{0.05916}{2}\log\dfrac{1}{a_\mathrm{Cu^{2+}}}\right) – \left(E^\circ_\mathrm{H^+/H_2} – \dfrac{0.05916}{2}\log\dfrac{f_\mathrm{H_2}}{(a_\mathrm{H^+})^2}\right)\\
&= \mathrm{\left(0.3419\: V - \dfrac{0.05916}{2}\log\dfrac{1}{0.0500}\right) – \left(0.0000 - \dfrac{0.05916}{2}\log\dfrac{0.500}{(0.100)^2}\right)}\\
&= \mathrm{+ 0.2531\: V}
\end{align}\]

 

Click here to return to the chapter.

Practice Exercise 11.3

Making appropriate substitutions into equation 11.3

\[\mathrm{+ 0.257\: V = \left(0.3419\: V - \dfrac{0.05916}{2}\log\dfrac{1}{\mathit{a}_\mathrm{Cu^{2+}}}\right) - \left(0.0000 - \dfrac{0.05916}{2}\log\dfrac{1.00}{(1.00)^2}\right)}\]

and solving for aCu2+ gives its activity as 1.35×10–3.

Click here to return to the chapter.

Practice Exercise 11.4

When using a saturated calomel electrode, the potential of the electrochemical cell is

\[E_\ce{cell} = E_\mathrm{UO_2^+/U^{4+}} - E_\ce{SCE}\]

Substituting in known values

\[\mathrm{−0.0190\: V = \mathit{E}_\mathrm{UO_2^+/U^{4+}} - 0.2444\: V}\]

and solving for EUO2+/U4+ gives its value as +0.2254 V. The potential relative to the Ag/AgCl electrode is

\[E_\ce{cell} = E_\mathrm{UO_2^+/U^{4+}} - E_\mathrm{Ag/AgCl} = \mathrm{0.2254\: V - 0.197\: V = +0.028\: V}\]

and the potential relative to the standard hydrogen electrode is

\[E_\ce{cell} = E_\mathrm{UO_2^+/U^{4+}} − E_\ce{SHE} = \mathrm{0.2254\: V − 0.0000\: V = +0.2254\: V}\]

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Practice Exercise 11.5

The larger the value of KA,I the more serious the interference. Larger values for KA,I correspond to more positive (less negative) values for logKA,I; thus, I, with a KA,I of 6.3×10–2, is the most serious of these interferents.

To find the activity of I that gives a potential equivalent to a NO2 activity of 2.75×10–4, we note that

\[a_\mathrm{NO_2^-} = K_\textrm{A.I} × a_\ce{I-}\]

Making appropriate substitutions

\[2.75 × 10^{−4} = (6.3×10^{−2}) × a_\mathrm{I^-}\]

and solving for aI gives its activity as 4.4×10–3.

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Practice Exercise 11.6

In the presence of OH the cell potential is

\[E_\ce{cell} = K - 0.05916\log\left\{a_\mathrm{NO_2^-} + K_\mathrm{NO_2^-/OH^−} × a_\mathrm{OH^-}\right\}\]

To achieve an error of less than 10%, the term KNO2/OH × aOH must be less than 1% of aNO2; thus

\[K_\mathrm{NO_2^−/OH^−} × a_\mathrm{OH^−} ≤ 0.10 × a_\mathrm{NO_2^−}\]

\[630 × a_\mathrm{OH^−} ≤ 0.10 × (2.2×10^{−4})\]

Solving for aOH gives its maximum allowable activity as 3.5×10–8, which corresponds to a pH of less than 6.54.

The electrode does have a lower pH limit. Nitrite is the conjugate weak base of HNO2, a species to which the ISE does not respond. As shown by the ladder diagram in Figure 11.58, at a pH of 4.15 approximately 10% of nitrite is present as HNO2. A minimum pH of 4.5 is the usual recommendation when using a nitrite ISE. This corresponds to an NO2/HNO2 ratio of

\[\mathrm{pH = p\mathit{K}_a + \log\dfrac{[\ce{NO2-}]}{[HNO_2]}}\]

\[\mathrm{4.5 = 3.15 + \log\dfrac{[\ce{NO2-}]}{[HNO_2]}}\]

\[\ce{\dfrac{[NO2- ]}{[HNO2]} ≈ 22}\]

Thus, at a pH of 4.5 approximately 96% of nitrite is present as NO2.

Figure11.58.jpg

Figure 11.58 Ladder diagram for the weak base NO2.

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Practice Exercise 11.7

The reduction of Cu2+ to Cu requires two electrons per mole of Cu (n = 2). Using equation 11.25, we calculate the moles and the grams of Cu in the portion of sample being analyzed.

\[N_\ce{Cu}= \dfrac{Q}{nF} = \mathrm{\dfrac{16.11\: C}{\dfrac{2\: mol\: \mathit{e}^−}{mol\: Cu} × \dfrac{96487\: C}{mol\: \mathit{e}^−}} = 8.348×10^{−5}\: mol\: Cu}\]

\[\mathrm{8.348×10^{−5}\: mol\: Cu × \dfrac{63.55\: g\: Cu}{mol\: Cu} = 5.301×10^{−3}\: g\: Cu}\]

This is the Cu from a 10.00 mL portion of a 500.0 mL sample; thus, the %/w/w copper in the original sample of brass is

\[\mathrm{\dfrac{5.301×10^{−3}\: g\: Cu × \dfrac{500.0\: mL}{10.00\: mL}}{0.442\: g\: sample} × 100 = 60.0\%\: w/w\: Cu}\]

For lead, we follow the same process; thus

\[N_\ce{Pb}= \dfrac{Q}{nF}= \mathrm{\dfrac{0.422\: C}{\dfrac{2\: mol\: \mathit{e}^-}{mol\: Pb} × \dfrac{96487\: C}{mol\: \mathit{e}^-}} = 2.19×10^{-6}\: mol\: Pb}\]

\[\mathrm{2.19×10^{−6}\: mol\: Pb × \dfrac{207.2\: g\: Pb}{mol\: Pb} = 4.53×10^{−4}\: g\: Pb}\]

\[\mathrm{\dfrac{4.53×10^{−4}\: g\: Pb × \dfrac{500.0\: mL}{10.00\: mL}}{0.442\: g\: sample} × 100 = 5.12\%\: w/w\: Pb}\]

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Practice Exercise 11.8

For anodic stripping voltammetry, the peak current, ip, is a linear function of the analyte’s concentration

\[i_\ce{p} = KC_\ce{Cu}\]

where K is a constant that accounts for experimental parameters such as the electrode’s area, the diffusion coefficient for Cu2+, the deposition time, and the rate of stirring. For the analysis of the sample before the standard addition we know that the current is

\[i_\ce{p} = 0.886 \hspace{20px} \ce{A} = KC_\ce{Cu}\]

and after the standard addition the current is

\[i_\ce{p} = \mathrm{2.52\: \mu A = \mathit{K}\left\{\mathit{C}_{Cu} × \dfrac{50.00\: mL}{50.005\: mL} + \dfrac{10.00\: mg\: Cu}{L} × \dfrac{0.005\: mL}{50.005\: mL}\right\}}\]

where 50.005 mL is the total volume after adding the 5.00 μL spike. Solving each equation for K and combining leaves us with the following equation.

\[\mathrm{\dfrac{0.886\: \mu A}{\mathit{C}_{Cu}} = \mathit{K} = \dfrac{2.52\: A}{\mathit{C}_{Cu} × \dfrac{50.00\: mL}{50.005\: mL} + \dfrac{10.00\: mg\: Cu}{L} × \dfrac{0.005\: mL}{50.005\: mL}}}\]

Solving this equation for CCu gives its value as 5.42×10-4 mg Cu2+/L, or 0.542 μg Cu2+/L.

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Practice Exercise 11.9

From the three half-wave potentials we have a ∆E1/2 of –0.280 V for 0.115 M en and a ∆E1/2 of –0.308 V for 0.231 M en. Using equation 11.51 we write the following two equations.

\[−0.280 = −\dfrac{0.05916}{2}\logβ_p − \dfrac{0.05916p}{2}\log(0.115)\]

\[−0.308 = −\dfrac{0.05916}{2}\log β_p − \dfrac{0.05916p}{2}\log(0.231)\]

To solve for the value of p, we first subtract the second equation from the first equation

\[0.028 = −\dfrac{0.05916p}{2}\log(0.115) − \left\{−\dfrac{0.05916p}{2}\log(0.231)\right\}\]

which eliminates the term with βp. Next we solve this equation for p

\[0.028 = (2.778×10^{-2})p - (1.882×10^{-2})p\]

\[0.028 = (8.96×10^{−3})p\]

obtaining a value of 3.1, or p ≈ 3. Thus, the complex is Cd(en)3. To find the formation complex, β3, we return to equation 11.51, using our value for p. Using the data for an en concentration of 0.115 M

\[−0.280 = −\dfrac{0.05916}{2}\logβ_p - \dfrac{0.05916 × 3}{2}\log(0.115)\]

\[−0.363 = −{0.05916}{2}\log β_p\]

gives a value for β3 of 1.93 × 1012. Using the data for an en concentration of 0.231 M gives a value of 2.10 × 1012.

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